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PostPosted: Thu May 08, 2014 2:01 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Checkerboard 8 NC DONC X 1

This is 8*8 and the Octets are a checkerboard Jigsaw.

It is X.
It is Non-Consecutive.

The cages are NC and Ordered increasing (standard Killer description order). They are also digitised in that one of the cell values is the last digit of the total.

Quite hard


Image

Solution:

17536248
84263571
48625317
71352684
35718426
62481753
26847135
53174862


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PostPosted: Thu May 15, 2014 2:36 am 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Back from vacation and saw this! Thanks HATMAN for the neat variant :applause:

I solved it using a couple of short chains... Took me much longer than it should have done because I forgot that the whole puzzle is NC - not just the cages.
I might try it in different orientations. See if there are unique solutions there as well :)

Clarifications and removal of inconsistencies thanks to Andrew.
Hidden Text:
0. The restrictions mean that the cages (all 3-cell) must be from:

Ending in 6 - [146], [246]
Ending in 7 - [137], [357]
Ending in 8 - [248], [258], [268], [468]

1. Cages must end in one of (678)
-> in D/ (678) locked in r1c8, r2c7, r7c2

2. Consider cages starting at r1c1, r3c4, r4c2.
Their end cells are all buddies and must form group (678).
Cage ending in 6 must have as middle cell 4.
-> One of r2c2, r4c3, r4c4 = 4
-> 4 nowhere else in Pink octet
Also none of those cages can be [248] or [468]!(Since r4c2 is also buddy with all possible locations for the 4).

3. At least one of r3c3 and r5c3 is from (67)
-> r4c3 cannot = 6 (NC)
-> r4c3 from (345)
-> r4c4 cannot be 4 (NC)
-> r5c5 cannot be 6 - must be from (78)
-> one of r35c3 = 6
-> r4c3 cannot be 5 (NC) - must be from (34)
-> r4c4 from (56) (NC)

4. r4c3 from (34) and r4c4 from (56) prevents cage starting at r3c4 from ending in 7. (I.e., cage cannot be [137] or [357]).
-> r5c5 = 8
-> r3c4 = 2
Also r35c3 = {67}

5. r3c4 = 2 and r5c3 from (67) prevents cage at r5c4 ending in 8.
-> r7c2 from (67) and r5c4 from (13)
But 3 in r5c4 puts 5 in r6c3 and 7 in r7c2 which leaves no solution for r5c3 (from 67).
-> r5c4 = 1

6. 8 in D/ in r1c8 or r2c7.
Also 8 in Pink in r1c2 or r2c1
-> 8 in Orange in r34c12
-> HS 8 in Grey -> r6c4 = 8

7. r3456c3, r4c4, and r7c2 are either

[6374][5][6] or [7463][6][7]

but the latter puts (245) (in Grey octet) in r78c1 and r8c2 -> must have 2 in r8c1 (NC)
which leaves no value for r4c5!

-> r3456c3, r4c4, and r7c2 = [6374][5][6]

Rest is clean up


Final Position. CPFC 11 th! P38 W 13 D 6 L 19. Pts 45. Prize money £73,207,049. WooHoo!


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PostPosted: Tue Jun 24, 2014 7:25 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another nice puzzle!

I only came across it at the weekend; I don't regularly visit this forum unless I'm working on the TJK Archive.

wellbeback wrote:
I solved it using a couple of short chains... Took me much longer than it should have done because I forgot that the whole puzzle is NC - not just the cages.
I loved wellbeback's step 2, which got into the puzzle more quickly than I did. Congratulations for managing to solve this puzzle without using NC for the whole grid. When that's used, it only seems necessary to use one chain.

I used a forcing chain, which gave the same result as wellbeback's first chain, locking cages and a lot of NC.

Here is my walkthrough for Chequerboard 8 NC DONC X 1:
Chequerboard octets are described by their colours, as in HATMAN’s diagram. Non-consecutive (NC) horizontally and vertically in cage pattern. Cages are Ordered (increasing), Non-consecutive, Digitised (last digit of cage total must be same as one of the cell values).

All cages must end with 6,7 or 8; cannot end with 5 because {135} doesn’t meet the Digitised condition.
Valid permutations ending in 6 are [146], [246]
Valid permutations ending in 7 are [137], [357]
Valid permutations ending in 8 are [248], [258], [268], [468]
-> first cells of cages must be {1234}, second cells {3456} and third cells {678}

1. Naked triple {678} in R1C8 + R2C7 + R7C2, locked for D/
1a. R6C3 = {345} -> no 4 in R6C24 + R7C3 (NC)

2. R1C8 + R2C7 = {678} -> no 6,7,8 in R1C7 + R2C8 (NC)
2a. No 6 in R1C7 -> cage at R1C6 cannot be {468}, no 4 in R1C6
2b. R1C6 = {123} -> no 2 in R1C5 (NC)
2c. R2C7 = {678} -> no 7 in R3C7 (NC)

3. R35C3 = {678} -> no 6 in R4C3 (NC)
3a. No 6 in R4C3 -> cage at R4C2 cannot be {468}, no 4 in R4C2
3b. R4C2 = {123} -> no 2 in R35C2 + R4C1 (NC)
3c. R4C3 = {345} -> no 4 in R4C4 (NC)

4. No 4 in R4C4 -> cage at R4C3 cannot be [146/246/248], no 6 in R5C5

5. R35C3 + R5C5 “see” each other so form naked triple {678}, 6 locked for C3
5a. 6 in R35C3 -> one of the cages at R1C1 and R4C2 must be [146/246] (locking cages) -> 4 in R2C2 + R4C3, locked for pink octet, no 4 in R1C12 + R2C1 + R3C4

6. R3C3 = {678} -> no 7 in R2C3 + R3C2 (NC)
6a. R5C3 = {678} -> no 7 in R5C2 (NC)
6b. R5C5 = {78} -> no 7,8 in R5C6 + R6C5 (NC)
6c. R7C2 = {678} -> no 7 in R6C2 + R7C13 + R8C2 (NC)

7. 7 in pink octet only in R1C2 + R2C1 + R3C3 -> no 6 in R2C2 (NC, using cage at R1C1)
7a. R1C1 = {123} -> no 2 in R1C2 + R2C1 (NC)
7b. R2C2 = {345} -> no 4 in R2C3 + R3C2 (NC)

8. Consider placements for 4 in pink octet
R2C2 = 4 => cage at R4C2 not [146/246], no 6 => R3C3 = 6 (hidden single in C3), R4C3 = 3 (not 5, NC) => R4C4 = 5 => cage at R3C4 = [258] (only remaining combination)
or R4C3 = 4 => R4C4 = 6 (not 3,5, NC) => cage at R3C4 = [268] (only remaining combination)
-> R3C4 = 2, placed for pink octet, no 2 in R1C1, R5C5 = 8, placed for D\ and white octet, no 8 in R7C8 + R8C7, R4C4 = {56}
8a. Naked pair {67} in R35C3, locked for C3
8b. R3C4 = 2 -> no 1,3 in R2C4 + R3C5 (NC)
8c. R4C4 = {56} -> no 5 in R4C35 (NC)
8d. R4C3 = {34} -> no 3 in R4C2 (NC)
8e. R4C2 = {12} -> no 1 in R35C2 + R4C1 (NC)

9. R35C3 = {67} -> one of the cages at R1C1 and R4C2 must contain 7 = [137/357] (locking cages) -> 3 in R1C1 + R2C2 + R4C3, locked for pink octet, no 3 in R1C2 + R2C1

10. Cage at R5C4 = [146/137/357], no 4 in R5C4, no 8 in R7C2
10a. 8 on D/ only in R1C8 + R2C7, locked for green octet, no 8 in R4C6
10b. Naked pair {67} in R5C3 + R7C2, locked for grey octet, no 6,7 in R6C4 + R7C1 + R8C2
10c. Naked pair {67} in R5C3 + R7C2, CPE no 6 in R5C2
10d. R7C2 = {67} -> no 6 in R6C2 (NC)

11. R5C3 + R7C2 = {67} -> one of the cages at R4C2 and R5C4 must contain 4 (locking cages), 4 in R46C3, locked for C3
11a. 4 in R46C3, CPE no 4 in R4C5 using D/

12. 8 in pink and green octets only in R1C2 + R2C1 and R1C8 + R2C7, X-Wing no other 8 in R12
12a. 8 in C3 only in R78C3, locked for blue octet, no 8 in R6C12 + R78C4
12b. R6C4 = 8 (hidden single in C4), placed for grey octet, no 8 in R7C1 + R8C2
12c. R6C4 = 8 -> no 7 in R7C4 (NC)

13. R137C2 = {678} (hidden triple in C2)
13a. R3C3 = {67} -> R3C2 = 8 (NC) -> R3C3 = 6 (NC), R4C4 = 5, both placed for D\, R4C3 = 3 (NC), placed for pink octet, R4C2 = 1 (NC), placed for orange octet, R1C1 = 1, R2C2 = 4, both placed for D\, R1C2 = 7, R2C1 = 8 (hidden pair in pink octet), R5C3 = 7
13b. R4C5 = 2, placed for D/ and green octet, no 2 in R2C8
13c. R2C1 = 8 -> no 7 in R3C1 (NC)
[Cracked. The rest is fairly straightforward.]

14. R7C2 = 6 -> cage at R5C4 = [146], all placed for D/ and grey octet, no 4 in R7C1
14a. R6C3 = 4 -> no 3,5 in R6C2 (NC) -> R6C2 = 2, placed for blue octet, no 2 in R5C1 + R78C3
14b. R6C2 = 2 -> no 3 in R5C2 + R6C1 (NC) -> R5C2 = 5, placed for blue octet, no 5 in R6C1 + R78C3
14c. R5C2 = 5 -> R5C1 = 3 (NC), placed for blue octet, no 3 in R78C4 -> R7C4 = 4
14d. R8C12 = [53], R4C1 = 4, R7C1 = 2

15. R2C7 = 7, placed for green octet -> R3C56 = [53], placed for green octet -> R1C7 = 4, placed for green octet -> R4C6 = 6, R4C178 = [784]

16. R6C6 = 7 -> no 6 in R6C57, no 8 in R7C6 (NC)

and the rest is naked singles, without using octets, diagonals and NC.


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